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August 2022 1 9 Report

СРОЧНО!!
1)[tex]( \frac{1}{1-m} - \frac{1}{1+m} -1) *( m^{2} -1)[/tex]

2)[tex]( \frac{m}{ n^{2}+mn } - \frac{m-n}{ m^{2} +mn }) : ( \frac{ n^{2} }{ m^{3} -mn^{2} } + \frac{1}{m+n} )[/tex]

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NNNLLL54

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$1)\; \; ( \frac{1}{1-m} - \frac{1}{1+m} -1)\cdot (m^2-1)= \frac{1+m-(1-m)-(1+m)(1-m)}{(1-m)(1+m)}\cdot (m^2-1)=\\\\= \frac{2m-(1-m^2)}{(1-m^2)} \cdot (m^2-1)= \frac{m^2+2m-1}{-(m^2-1)}\cdot (m^2-1) =1-2m-m^2$

$2)\; \; ( \frac{m}{n^2+mn} - \frac{m-n}{m^2+mn} ):( \frac{n^2}{m^3-mn^2} +\frac{1}{m+n} )=\\\\=( \frac{m}{n(m+n)} -\frac{m-n}{m(m+n)} ):( \frac{n^2}{m(m^2-n^2)}+\frac{1}{m+n})=\\\\= \frac{m^2-n(m-n)}{mn(m+n)} : \frac{n^2+m(m-n)}{m(m-n)(m+n)} = \frac{m^2-mn+n^2}{mn(m+n)} \cdot \frac{mn(m-n)(m+n)}{n^2-mn+m^2} =\\\\=m-n$

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